Board Previous Year Questions

How Boards Test This Chapter

Dual Nature is a compact, high-certainty scorer in CBSE papers. The recurring demands:

• 1 mark: define work function / threshold frequency / stopping potential; state the effect of intensity or frequency on photocurrent or $K_{max}$; write the de Broglie relation.
• 2 marks: graphs — sketch photocurrent vs potential for two intensities or two frequencies; $V_0$ vs $\nu$ and what its slope/intercepts give; short numericals on $K_{max} = h\nu - \phi_0$.
• 3 marks: state Einstein's photoelectric equation and use it to explain the experimental laws; why the wave theory fails; full numericals chaining photon energy → $K_{max}$ → $V_0$ → $v_{max}$; de Broglie wavelength of accelerated electrons.
• 5 marks: describe the experimental study of the photoelectric effect with diagrams and all three graphs, then derive/interpret via Einstein's equation.

The questions below are Board-style previous-year questions with full step-by-step solutions embedded in the explanations. Attempt each before reading its solution.

Where a specific exam-year attribution is certain it is included; otherwise questions are tagged simply [CBSE Board] to avoid misattribution.

The Definitions Boards Reward (Model Answers)

Work function: the minimum energy required by an electron to escape from the metal surface. Depends on the metal and the nature of its surface. Unit: eV.

Threshold frequency: the minimum frequency of incident radiation below which no photoelectric emission takes place from a given photosensitive material, no matter how intense the light. $\nu_0 = \phi_0/h$.

Stopping (cut-off) potential: the minimum negative (retarding) potential given to the collector plate for which the photocurrent becomes zero. $eV_0 = K_{max}$. For a given frequency it is independent of intensity.

Einstein's photoelectric equation: $K_{max} = h\nu - \phi_0$, based on: radiation consists of quanta of energy $h\nu$; photoelectric emission results from the absorption of a single quantum by a single electron; energy conservation applied to that elementary event.

de Broglie relation: $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$ — a moving particle of momentum p has an associated matter wave of wavelength $\lambda$. The relation holds for photons too.

[Board Tip] Marks are lost most often for (i) forgetting 'minimum' in the definitions, (ii) drawing I-V curves that don't saturate, (iii) $V_0$-$\nu$ lines drawn through the origin — the intercept at $\nu_0$ is the physics.

Board PYQ Set A: Assertions, Definitions & Graphs (1-2 marks)

PYQ 1. Define the term 'stopping potential' in relation to the photoelectric effect. [CBSE Board]

Solution: The minimum retarding (negative) potential applied to the collector plate at which the photoelectric current becomes zero is called the stopping potential $V_0$. It measures the maximum kinetic energy of emitted photoelectrons: $K_{max} = eV_0$.

PYQ 2. The stopping potential in an experiment is found to be independent of the intensity of incident radiation. What information does this give about the emitted photoelectrons? [CBSE Board]

Solution: Since $eV_0 = K_{max}$, the maximum kinetic energy of the photoelectrons is independent of intensity. Increasing intensity increases only the NUMBER emitted per second (saturation current), not their energies — direct support for the one-photon-one-electron picture.

PYQ 3. Draw the variation of photocurrent with collector potential for a fixed frequency but two intensities $I_2 > I_1$. [CBSE Board]

Solution: Two curves rising from the same stopping potential $-V_0$ on the negative axis (same frequency → same $V_0$), each saturating on the positive side; the $I_2$ curve saturates at double-height if $I_2 = 2I_1$. Key labelled features: common $-V_0$, distinct saturation levels proportional to intensity.

PYQ 4. Show graphically how the stopping potential varies with the frequency of incident radiation for two metals A and B having work functions $\phi_A > \phi_B$. What does the slope represent? [CBSE Board]

Solution:

1. Both are straight lines with the SAME slope $h/e$ (universal), meeting the frequency axis at $\nu_{0A} > \nu_{0B}$ (since $\nu_0 = \phi_0/h$).
2. The lines are parallel; A's line sits to the right (higher threshold) with a more negative y-intercept ($-\phi_A/e$).
3. Slope = $h/e$ — Planck's constant divided by electronic charge.

PYQ 5. Red light, however bright, cannot eject photoelectrons from a zinc surface, but even weak ultraviolet light can. Explain on the basis of photon theory. [CBSE Board]

Solution:

1. Emission requires each absorbed photon to supply at least the work function: $h\nu \geq \phi_0$.
2. Red photons: $h\nu \approx$ 1.8-2 eV < $\phi_0$(zinc); brightness only adds MORE inadequate photons — no single one suffices, and electrons absorb quanta singly.
3. UV photons individually exceed $\phi_0$, so even a few (weak light) eject electrons immediately.

PYQ 6. Why is the wave theory unable to explain the instantaneous nature of photoelectric emission? [CBSE Board]

Solution:

1. In the wave picture, energy is spread continuously over the wavefront and shared by an enormous number of electrons.
2. The energy absorbed per electron per unit time is then minuscule; accumulating a few eV would take hours.
3. Observed emission occurs within ~$10^{-9}$ s at any intensity — possible only if energy arrives in concentrated quanta absorbed whole.

Board PYQ Set B: Standard Numericals (2-3 marks)

PYQ 7. The work function of caesium is 2.14 eV. Find (a) the threshold frequency, and (b) the wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. [CBSE Board]

Solution:

1. (a) $\nu_0 = \phi_0/h = \dfrac{2.14 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.16 \times 10^{14}$ Hz.
2. (b) $\lambda = \dfrac{hc}{eV_0 + \phi_0} = \dfrac{1240 \text{ eV nm}}{2.74 \text{ eV}} \approx 454$ nm.

PYQ 8. Light of frequency $7.21 \times 10^{14}$ Hz on a metal surface ejects electrons of maximum speed $6.0 \times 10^5$ m/s. Calculate the threshold frequency. [CBSE Board]

Solution:

1. $K_{max} = \frac{1}{2}mv^2 = \frac{1}{2}(9.1 \times 10^{-31})(3.6 \times 10^{11}) = 1.64 \times 10^{-19}$ J.
2. $\nu_0 = \nu - K_{max}/h = 7.21 \times 10^{14} - 2.47 \times 10^{14} = 4.74 \times 10^{14}$ Hz.

PYQ 9. The threshold frequency of a metal is $3.3 \times 10^{14}$ Hz. Light of frequency $8.2 \times 10^{14}$ Hz is incident on it. Calculate the cut-off voltage. [CBSE Board]

Solution:

1. $eV_0 = h(\nu - \nu_0) = 6.63 \times 10^{-34} \times 4.9 \times 10^{14} = 3.25 \times 10^{-19}$ J.
2. $V_0 = \dfrac{3.25 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.0$ V.

PYQ 10. An electron and a proton have the same kinetic energy. Which has the greater de Broglie wavelength? Justify. [CBSE Board]

Solution:

1. $\lambda = \dfrac{h}{\sqrt{2mK}}$; at equal K, $\lambda \propto \dfrac{1}{\sqrt{m}}$.
2. The electron, being ~1836 times lighter, has the greater wavelength ($\lambda_e/\lambda_p = \sqrt{1836} \approx 43$).

PYQ 11. Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V. [CBSE Board]

Solution:

1. $\lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{1.227}{\sqrt{V}}$ nm.
2. $\lambda = \dfrac{1.227}{10} = 0.123$ nm.

PYQ 12. A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths. [CBSE Board]

Solution:

1. $\lambda = \dfrac{h}{\sqrt{2mqV}} \propto \dfrac{1}{\sqrt{mq}}$.
2. $\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4 \times 2} = 2\sqrt{2}$.
3. $\lambda_p : \lambda_\alpha = 2\sqrt{2} : 1$.

PYQ 13. Monochromatic light of frequency $6.0 \times 10^{14}$ Hz is produced by a laser emitting $2.0 \times 10^{-3}$ W. Find the energy of each photon and the number emitted per second. [CBSE Board]

Solution:

1. $E = h\nu = (6.63 \times 10^{-34})(6.0 \times 10^{14}) = 3.98 \times 10^{-19}$ J.
2. $N = P/E = \dfrac{2.0 \times 10^{-3}}{3.98 \times 10^{-19}} = 5.0 \times 10^{15}$ photons per second.

Board PYQ Set C: Long-Answer Patterns (3-5 marks)

PYQ 14. State Einstein's photoelectric equation. Explain how it accounts for (i) the independence of $K_{max}$ from intensity, (ii) the existence of a threshold frequency, and (iii) the instantaneous nature of emission. [CBSE Board]

Solution:

1. $K_{max} = h\nu - \phi_0$: one electron absorbs one quantum $h\nu$; after paying the work function, the rest is kinetic energy.
2. (i) $K_{max}$ contains only $\nu$ and $\phi_0$; intensity (photon count per second) never enters the per-electron energy budget.
3. (ii) $K_{max} \geq 0$ demands $\nu \geq \phi_0/h = \nu_0$; below it, no photon can free an electron, at any intensity.
4. (iii) absorption of a quantum is a single instantaneous elementary event — no accumulation time; dim light only reduces the event rate.

PYQ 15. With a neat labelled diagram, describe the experimental arrangement for studying the photoelectric effect. Sketch the variation of photocurrent with (a) intensity, (b) collector potential at different intensities, (c) collector potential at different frequencies. [CBSE Board]

Solution (outline the examiner expects):

1. Diagram: evacuated glass/quartz tube; photosensitive emitter C; collector A; quartz window; monochromatic source; commutator-reversible battery; voltmeter; microammeter.
2. (a) photocurrent vs intensity: straight line through the origin.
3. (b) I-V at intensities $I_3 > I_2 > I_1$ (same $\nu$): all cut off at the same $-V_0$; saturation levels increase with intensity.
4. (c) I-V at frequencies $\nu_3 > \nu_2 > \nu_1$ (same intensity): same saturation level; cut-offs at increasingly negative potentials $V_{03} > V_{02} > V_{01}$.

PYQ 16. (a) Why is the wave theory of light unable to explain the photoelectric effect? Give three reasons. (b) How does Einstein's photon picture resolve each? [CBSE Board]

Solution:

1. Wave theory predicts $K_{max}$ grows with intensity — observed: intensity-independent. Photon fix: energy per absorption event is $h\nu$, set by frequency alone.
2. Wave theory predicts no threshold — observed: sharp $\nu_0$. Photon fix: single photon must beat $\phi_0$, so $\nu \geq \phi_0/h$.
3. Wave theory predicts hours-long delays in dim light — observed: emission in ~$10^{-9}$ s. Photon fix: absorption is one instantaneous elementary process.

PYQ 17. Derive an expression for the de Broglie wavelength of an electron accelerated through a potential difference V. Hence show $\lambda \approx \dfrac{1.227}{\sqrt{V}}$ nm. [CBSE Board]

Solution:

1. Kinetic energy gained: $K = eV$; momentum $p = \sqrt{2mK} = \sqrt{2meV}$.
2. de Broglie: $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2meV}}$.
3. Substituting h = $6.63 \times 10^{-34}$ J s, m = $9.1 \times 10^{-31}$ kg, e = $1.6 \times 10^{-19}$ C: $\lambda = \dfrac{1.227 \times 10^{-9}}{\sqrt{V}}$ m $= \dfrac{1.227}{\sqrt{V}}$ nm.